Attack log — a smaller constant in the two-squares gap

2026-08-17 — literature (do not claim Landau–Ramanujan)

Fetched, in order:

• Green, 100 Open Problems (Dec 2025 update), Problem 66: “Is there always a sum of two squares between $X-\tfrac1{10}X^{1/4}$ and $X$?” Comments: Littlewood’s list (Problem 2), Montgomery’s list, and Erdős, Some unsolved problems, Michigan Math. J. 4 (1957), Problem 15. Elementary $O(X^{1/4})$ by subtracting two successive largest squares. Green–Lindqvist give two almost equal squares in the same $O(X^{1/4})$ scale. No later update on the constant.
• The proposed-50 pointer “Erdős Problem #143” is a numbering error. Bloom #143 is the primitive-set sparsity problem. The matching Bloom page is Erdős Problem #222 (accessed 2026-08-17; cites [Er57] and [Er61]). Status: OPEN, “cannot be resolved with a finite computation”. Zero comments, zero claimed proofs.
• Erdős 1957, Problem 15 (fetched the Michigan Math. J. scan): if $(a_k)$ is the increasing sequence of sums of two squares, Bambah– Chowla give $a_{k+1}-a_k=O(a_k^{1/4})$; “it has not yet been shown that the $O$ can be replaced by $o$”. Lower bound $a_{k+1}-a_k>c(\log a_k)/\log\log a_k$ infinitely often.
• Bambah–Chowla, Proc. Nat. Inst. Sci. India 13 (1947): if $\alpha>2\sqrt{2}$ then for all large $n$ there are $u,v$ with $n\le u^2+v^2<n+\alpha n^{1/4}$. Arithmetic form (Shiu 2013/2019): $u=\lfloor\sqrt n\rfloor$, $v=\lceil\sqrt{n-u^2}\rceil$ gives $n\le u^2+v^2<n+2\sqrt{2}\,n^{1/4}+1$ for every $n\ge 1$.
• Uchiyama 1964: the $+1$ may be replaced by $-1$.
• Jameson, Math. Gazette 103 (2019): replaced by $-2$.
• Shiu, Integers 19 (2019), #A48: even $a=3$ (i.e. $n\le u^2+v^2<n+2\sqrt{2}\,n^{1/4}-3$) is open, and he produces an infinite family of $n$ on which both the Bambah–Chowla point and $(u+1)^2$ fail $a=3$. Under a fractional-part hypothesis on $\sqrt{2up+v^2-p^2}$ he would get every $\alpha>0$.
• Richards 1982: $\limsup(s_{n+1}-s_n)/\log s_n\ge 1/4$.
• Dietmann–Elsholtz–Kalmynin–Konyagin–Maynard, IMRN 2023 (arXiv:1810.03203): the log-scale constant is at least $195/449\approx 0.434$ in the arXiv v2 writeup; Bloom #222 quotes the published improvement $0.868\cdots$. Logarithmic gaps do not move the $X^{1/4}$ coefficient. Isolated gap tables are residue unless they force a new infinite-family bound at the $X^{1/4}$ scale.
• Guo–Ilyin, J. Number Theory 2024: empty annuli of width $\lambda^s$ for $s<1/4$; at $s=1/4$ the Bambah–Chowla scale already forces a lattice point. No new coefficient.
• Iyer, arXiv:2503.15789 (2025): generalises Bambah–Chowla to $x^\theta+y^\theta$; for $\theta=2$ restates $2\sqrt{2}\,x^{1/4}+1$. No smaller constant.

Published record I must beat, after the fetch:

Here $G(n)=\min\{s-n:s\ge n,\ s=a^2+b^2\}$. Backward gaps $n-s$ are the same sequence of consecutive differences, so Green’s interval $[X-\tfrac1{10}X^{1/4},X]$ is the same constant.

Do not claim the Landau–Ramanujan density. Average-gap heuristics are not a bound.

2026-08-17 — tonight’s handle

Three live handles, in order of what would count as a dent:

1. Beat the published multiplicative constant. A proof that $G(n)<\alpha n^{1/4}$ for some explicit $\alpha<2\sqrt{2}$ and all large $n$. Shiu calls this open for seventy-plus years.
2. Beat Jameson’s additive $-2$. Prove $a=3$: $G(n)<2\sqrt{2}\,n^{1/4}-3$ for all large $n$. Shiu isolates the obstruction as an infinite 1-parameter family $n=u^2+m^2+1$ with $m$ even and $2u+1=(m+1)^2$. On that family both the BC-point and $(u+1)^2$ sit at $n+2m$, while $2m+2<\Phi_n<2m+3$, so $a=3$ is exactly the statement that $(u^2+m^2,\,u^2+(m+1)^2)$ contains another two-square.
3. Infinite-family lower bound at the $X^{1/4}$ scale. If that same interval is empty for infinitely many even $m$, then $\limsup G(n)/n^{1/4}=2\sqrt{2}$ and Bambah–Chowla is sharp; in particular Green’s $1/10$ fails for infinitely many $X$. Isolated empty intervals are residue unless the emptiness is infinite.

Starting with (2) and (3) on the Shiu family, then a certified replay of the elementary $2\sqrt{2}$ bound, then a gap census.

2026-08-17 — numbering correction, recorded

Erdős 1957 Problem 15 = Bloom #222, not #143. Green #66 is the explicit-constant form of the same question. Work continues under the house folder two-squares-gap.

2026-08-17 — elementary replay

verify_bc.py --N 100000: Bambah–Chowla leftover never meets $\Phi+1$. Two-point $a=2$ never fails. Two-point $a=3$ fails on 172 values; the sample is ladder tops $1,3,6,21,30,59,\dots,161$ (Shiu shape).

2026-08-17 — Shiu family is occupied

Even $m\le 8000$, family $2u+1=(m+1)^2$, $n=u^2+m^2+1$: the open interval $(u^2+m^2,u^2+(m+1)^2)$ is empty only for $m=2$ ($n=21$). Occupied thereafter. Least $p=u-u'$ that saves the interval has maximum $40$, at Shiu’s example $m=2862$. No infinite-family sharpness of $2\sqrt{2}$ from this curve.

2026-08-17 — third-point hunt

hunt_a3.py on $n\le 2\cdot 10^5$: 239 two-point $a=3$ failures, 235 saved, 4 unsaved: 3, 6, 21, 91. Ladder enumerator through $u=20000$ ($n\le 4\cdot 10^8$): still those four. Every saved failure has rung $t=0$.

2026-08-17 — integer classification

Two-point $a=3$ fails $\Leftrightarrow(\min(h,h_2)+3)^4\ge 64n$. Expansions: rungs $t\ge 1$ never fail; $k\le 2m-3$ and $k\ge 3m+3$ never fail. Obstruction = ladder tops with $k\in[2m-2,3m+2]$. And $m\le\sqrt{2}\,n^{1/4}$.

2026-08-17 — danger-zone search

certify_a3.py:

Stored witnesses for $M=250$ and $M=2000$ pass verify_a3_cert.py (rebuilds $n$ from $(m,k)$, checks $a^2+b^2=s$ and $(s-n+3)^4<64n$).

2026-08-17 — exhaustive table

verify_exhaustive.py --N 5000000: generate every two-square, set $G(n)=$ next one minus $n$. Failures of $a=3$: exactly $\{1,3,6,21,91\}$. Failures of $a=2$: none.

2026-08-17 — gap census (residue, not a bound)

Through $N=2\cdot 10^7$, max $(s_{k+1}-s_k)/s^{1/4}$ is $2.407$ at $1493\to 1508$. Green’s $1/10$ is still crossed at $X=2\cdot 10^7$. Not a large-$X$ counterexample scheme; Richards gaps are the wrong scale. Plot: figures/gap_ratios.png.

2026-08-17 — what we will claim

A certified improvement of Jameson’s published additive $-2$ to $-3$, for all $2\le n\le 1.024\cdot 10^{15}$ except $\{3,6,21,91\}$. Not Green’s $1/10$. Not a multiplicative $\alpha<2\sqrt{2}$. Not $a=3$ for every large $n$. Do not claim Landau–Ramanujan.